# Homework Help: % polarization of partially polarized light

1. Mar 14, 2009

### ken~flo

1. Here's the question:

The percent polarization P of a partially polarized beam of light is defined as

P = [Imax - Imin]/[Imax + Imin] x 100

where Imax and Imin are the max and min intensities that are obtained when the light
passes through a polarizer that is slowly rotated. Such light can be considered as the
sum of two unequal plane-polarized beams of intensities Imax and Imin perpindicular to
eachother. Show that the light transmitted by a polarizer, whose axis makes an angle
theta to the direction in which Imax is obtained, has intensity

[1 + pcos(2theta)]/[1 + p]

where p = P/100.

2. Relevant equations

I = I# x cos^2(theta) (I# is I naught, or initial intensity)

3. I thought that the intensity would be at a maximum when theta=0 degrees, because
cosine of 0 is 1, so the Intensity of the polarized light would be equal to the intensity
of the light initially, but that doesn't seem to work out to something that resembles
the answer. I also thought that maybe p=cos^2(theta), since it is the fractional
percentage of the initial intensity. I also tried coming up with equations for Imax and
Imin, but I wasn't sure whether or not they would have the same angles and
intensities. Any help with this problem would be greatly appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Mar 15, 2009
2. Mar 15, 2009

### Staff: Mentor

Remember that you should add the two contributions for the two orthogonal "sources". So write one equation for the intensity distribution of one source, and another for the intensity distribution of the orthogonal source, and then add them, under the constraint of the top equation for the percent variation between the two sources....

Show us some of those equations, and I think you will be mostly there...

3. Mar 15, 2009

### ken~flo

okay, so here's what I have so far

I (as a function of theta) = Imaxcos^2theta
Imaxcos^2theta = Imax + Imin
Imax(cos^2theta-1) = Imin
Imax(cos2theta) = Imin

am I on the right path?